Question: The equation of a circle $C$ is $x^2+y^2+4x+14y+28 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+4x) + (y^2+14y) = -28$ $(x^2+4x+4) + (y^2+14y+49) = -28 + 4 + 49$ $(x+2)^{2} + (y+7)^{2} = 25 = 5^2$ Thus, $(h, k) = (-2, -7)$ and $r = 5$.